C# - 189 Bytes
double f(double x, double y){double r,t,m,c;r=Math.Sqrt(x*x+y*y);t=Math.Atan2(y,x);m=Math.Pow(r,-x)*Math.Exp(y*t-t);c=Math.Cos((1-y)*Math.Log(r)-t*x);return m*(2*c*c<1?Math.Sqrt(1-c*c):c);}
Readable:
double f(double x, double y){
double r, t, m, c;
r = Math.Sqrt(x * x + y * y);
t = Math.Atan2(y, x);
m = Math.Pow(r, -x) * Math.Exp(y * t - t);
c = Math.Cos((1 - y) * Math.Log(r) - t * x);
return m * (2 * c * c < 1 ? Math.Sqrt(1 - c * c) : c); }
Explanation: Decided not to use any Complex libraries.
$$
\begin{align}
z & = x + iy \\
& = r e^{it} \\
z^{i-z} & = (re^{it})^{(-x +i(1-y))} \\
& = r^{-x} r^{i(1-y)} e^{-xit} e^{t(y-1)} \\
& = r^{-x} e^{t(y-1)} e^{i((1-y)\ln(r)-xt)} \text{ (as }r^i = e^{i\ln(r)}\text{)}
\end{align}
$$
Let this be equal to \$me^{ia}\$ where
$$m = r^{-x} e^{t(y-1)}$$
$$a = (1-y)\ln(r)-xt$$
Then \$\Re(z^{i-z})=m \cos a\$ and \$\Im(z^{i-z})=m \sin a\$
The maximum absolute value can be determined by the \$\cos a\$ and \$\sin a\$ terms, with these being equal at \$\frac{1}{\sqrt{2}}\$ (hence the test \$2c^2 < 1\$).
As mentioned, raising to a complex exponent is dependent on choosing a particular branch cut ( e.g. \$z = 1\$ could be \$e^{i\pi}\$ or \$e^{3i\pi}\$ - raising this to \$i\$ gives a real part of \$e^{-\pi}\$ or \$e^{-3\pi}\$ respectively), however, I have just used the convention of \$t \in [0,2\pi)\$ as per the question.
-2+icould be used for that (z^(i-z)=3-4iso3>-4vsabs(-4)>abs(3)). \$\endgroup\$